Central Extensions of Three Infinite-Dimensional Algebras

Derives Theorem A39.2. Let \(c\) be a \(2\)-cocycle on \(\mathfrak{w}\).

Step 1: normalize against \(L_{0}\). Define a linear functional \(b\) on \(\mathfrak{w}\) by

Its coboundary is \((\partial b)(L_{m},L_{n}) = b\left(\comm{L_{m}}{L_{n}}\right) = (m-n)\,b(L_{m+n})\), so for \(n \neq 0\)

while \((\partial b)(L_{0},L_{0}) = 0 = c(L_{0},L_{0})\) by antisymmetry. Replacing \(c\) by \(c - \partial b\) — which changes nothing in \(H^{2}\) — we may and do assume

Step 2: only the diagonal \(m+n = 0\) survives. Impose the cocycle condition Equation (18.121) on the triple \(\left(L_{0},L_{m},L_{n}\right)\). With \(\comm{L_{0}}{L_{m}} = -mL_{m}\), \(\comm{L_{n}}{L_{0}} = nL_{n}\) and \(\comm{L_{m}}{L_{n}} = (m-n)L_{m+n}\),

the middle term vanishing by Equation (A39.5) and the last by antisymmetry. Hence

and antisymmetry gives \(h(-m) = -h(m)\), with \(h(0) = 0\).

Step 3: the recursion. A cocycle of the shape Equation (A39.6) makes every term of Equation (18.121) vanish on a triple \(\left(L_{m},L_{n},L_{k}\right)\) with \(m+n+k \neq 0\), since each term pairs \(L_{m+n}\) with \(L_{k}\) and the two indices sum to \(m+n+k\). For \(m+n+k = 0\) the three terms are

so the cocycle condition is \((m-n)h(k) + (n-k)h(m) + (k-m)h(n) = 0\) with \(k = -m-n\), that is

Setting \(n = 1\),

which for \(m \ge 2\) determines \(h(m+1)\) from \(h(m)\) and \(h(1)\). Since \(h(2)\) is not reached by Equation (A39.8) — at \(m = 1\) the left-hand side vanishes identically, and indeed the relation then reads \(3h(1) = 3h(1)\) — the solution space has dimension at most \(2\), the free data being \(h(1)\) and \(h(2)\); negative arguments follow from \(h(-m) = -h(m)\).

Two solutions are exhibited directly. For \(h(m) = m\),

and for \(h(m) = m^{3}\), expanding both sides,

where the middle expression is obtained by multiplying out \((m-n)\left(m^{3}+3m^{2}n+3mn^{2}+n^{3}\right)\) and cancelling the two \(3m^{2}n^{2}\) terms. The two are linearly independent, so the solution space of Equation (A39.7) is exactly \(\set{h(m) = \alpha m + \beta m^{3}}\).

Step 4: the linear solution is a coboundary. Take \(b_{\gamma}\left(L_{n}\right) = \gamma\,\delta_{n,0}\). Then \(\left(\partial b_{\gamma}\right)(L_{m},L_{n}) = (m-n)\gamma\,\delta_{m+n,0} = 2m\gamma\,\delta_{m+n,0}\), which is Equation (A39.6) with \(h(m) = 2\gamma m\); and it respects Equation (A39.5), since \(\left(\partial b_{\gamma}\right)(L_{n},L_{0}) = n\,b_{\gamma}(L_{n}) = 0\) for \(n \neq 0\). Choosing \(\gamma = \alpha/2\) removes the \(\alpha\) term, so every class is represented by a multiple of \(h(m) = m^{3}\), and equally by a multiple of Equation (A39.2), since \((m^{3}-m)/12\) is such a combination. Thus \(\dim H^{2}(\mathfrak{w},F) \le 1\).

Step 5: \(\omega\) is not a coboundary. Every coboundary satisfies \(\left(\partial b\right)\left(L_{m},L_{-m}\right) = b\left(\comm{L_{m}}{L_{-m}}\right) = 2m\,b\left(L_{0}\right)\), a linear function of \(m\). If \(\omega = \partial b\) then \((m^{3}-m)/12 = 2m\,b(L_{0})\) for every \(m\); at \(m = 2\) this gives \(b(L_{0}) = \tfrac{1}{8}\) and at \(m = 3\) it gives \(b(L_{0}) = \tfrac{1}{3}\), a contradiction. Hence \(\dim H^{2}(\mathfrak{w},F) = 1\) and \(\omega\) generates it.

The last assertion is Proposition 18.76: extensions up to equivalence are in bijection with classes, the class of \(C\,\omega\) corresponds to the bracket Equation (A39.3) through Equation (18.120), and by Equation (18.123) the extension is trivial precisely for \(C = 0\).

Derives Lemma A39.5. \(\left(t^{n}\right)' = n\,t^{n-1}\), whose exponent is \(-1\) only for \(n = 0\), where the coefficient \(n\) vanishes; so no \(t^{-1}\) term is ever produced and \(\operatorname{Res}f' = 0\) by linearity. Applying this to \(fg\) and using the Leibniz rule, \(0 = \operatorname{Res}\left(fg\right)' = \operatorname{Res}\left(f'g\right) + \operatorname{Res}\left(fg'\right)\).

Derives Theorem A39.6. The two forms agree. For \(f = t^{m}\), \(g = t^{n}\) one has \(f'g = m\,t^{m+n-1}\), whose residue is \(m\,\delta_{m+n,0}\).

Antisymmetry. \(\kappa\) is symmetric (Definition 18.10), so

by Lemma A39.5.

The cocycle condition. Let \(u = x\otimes f\), \(v = y\otimes g\), \(w = z\otimes h\). By Equation (A39.9),

and likewise for the two cyclic permutations. The three Killing-form factors are equal. Indeed Lemma 18.11 reads \(\kappa\left(\comm{X}{Y},Z\right) = -\kappa\left(Y,\comm{X}{Z}\right)\); taking \(X = y\), \(Y = x\), \(Z = z\) and using antisymmetry of the bracket gives \(\kappa\left(\comm{x}{y},z\right) = \kappa\left(x,\comm{y}{z}\right)\), and the symmetry of \(\kappa\) turns the right side into \(\kappa\left(\comm{y}{z},x\right)\); applying the same step again gives \(\kappa\left(\comm{z}{x},y\right)\). Write \(\mathcal{K}\) for the common value. The cyclic sum is therefore

Expanding each derivative by the Leibniz rule,

and \(\operatorname{Res}\) of a derivative vanishes by Lemma A39.5. So Equation (A39.12) is zero, which is Equation (18.121).

It is not a coboundary. Since \(\mathfrak{g}\) is simple, \(\comm{\mathfrak{g}}{\mathfrak{g}}\) is a nonzero ideal, hence all of \(\mathfrak{g}\); so \(\kappa \neq 0\), and there is \(x \in \mathfrak{g}\) with \(\kappa(x,x) \neq 0\) — were \(\kappa(x,x) = 0\) for every \(x\), the polarization identity \(2\kappa(x,y) = \kappa(x+y,x+y) - \kappa(x,x) - \kappa(y,y)\) would make \(\kappa\) vanish identically, contradicting its nondegeneracy (Theorem 18.13). Put \(u = x\otimes t\) and \(v = x\otimes t^{-1}\). Then \(\comm{u}{v} = \comm{x}{x}\otimes 1 = 0\), so \(\left(\partial b\right)(u,v) = b\left(\comm{u}{v}\right) = 0\) for every linear \(b\), whereas \(\omega_{\kappa}(u,v) = 1\cdot\kappa(x,x) \neq 0\) by Equation (A39.10). The extension Equation (A39.11) is then Equation (18.120) for the cocycle \(\ell\,\omega_{\kappa}\), and is nontrivial for \(\ell \neq 0\) by Equation (18.123).

Derives Lemma A39.7. Give \(\mathfrak{g}^{*}\) the coadjoint action \(\left(z\cdot\phi\right)(y) = -\phi\left(\comm{z}{y}\right)\). For a bilinear \(B\) satisfying Equation (A39.13) the map \(\Phi_{B} : \mathfrak{g}\rightarrow\mathfrak{g}^{*}\), \(\Phi_{B}(x) = B(x,\cdot)\), obeys

so \(\Phi_{B}\) intertwines the adjoint and coadjoint actions. The Killing form satisfies Equation (A39.13) by Lemma 18.11 and is nondegenerate by Theorem 18.13, so \(\Phi_{\kappa}\) is an isomorphism of \(\mathfrak{g}\)-modules. Hence \(T = \Phi_{\kappa}^{-1}\circ\Phi_{B}\) is an endomorphism of \(\mathfrak{g}\) commuting with every \(\ad_{z}\). The adjoint representation of a simple algebra on the finite-dimensional complex space \(\mathfrak{g}\) is irreducible — an invariant subspace is an ideal — so Schur's lemma, in the form of Theorem 9.158 (whose proof uses only that \(T\) commutes with an irreducible family of operators on a finite-dimensional complex space), gives \(T = \lambda\,\identity\) and therefore \(B = \lambda\,\kappa\).

Derives Proposition A39.8. Write \(B_{m}(x,y) = c\left(x\otimes t^{m},y\otimes t^{-m}\right)\). Antisymmetry of \(c\) gives

so \(B_{0}\) is an antisymmetric bilinear form on \(\mathfrak{g}\otimes 1 \cong \mathfrak{g}\).

Step 1: kill \(B_{0}\). The restriction of a cocycle to a subalgebra is a cocycle, so \(B_{0}\) is a \(2\)-cocycle on \(\mathfrak{g}\); by Proposition 18.77, \(H^{2}(\mathfrak{g},\C) = 0\), so \(B_{0} = \partial\beta\) for a linear \(\beta\) on \(\mathfrak{g}\). Extend \(\beta\) to \(b\) on \(L\mathfrak{g}\) by \(b(x\otimes t^{0}) = \beta(x)\) and \(b(x\otimes t^{n}) = 0\) for \(n\neq0\). Then \(\left(\partial b\right)\left(x\otimes t^{m},y\otimes t^{n}\right) = \delta_{m+n,0}\,\beta\left(\comm{x}{y}\right)\), which is homogeneous of degree zero, so \(c - \partial b\) is still homogeneous of degree zero and now has \(B_{0} = 0\). Replace \(c\) by it.

Step 2: each \(B_{m}\) is invariant. Apply Equation (18.121) to \(\left(z\otimes t^{0},\ x\otimes t^{m},\ y\otimes t^{-m}\right)\):

and \(B_{0} = 0\) while \(B_{-m}\left(\comm{y}{z},x\right) = -B_{m}\left(x,\comm{y}{z}\right) = B_{m}\left(x,\comm{z}{y}\right)\) by Equation (A39.14). What is left is exactly Equation (A39.13), so \(B_{m} = \lambda_{m}\,\kappa\) by Lemma A39.7. Feeding that back into Equation (A39.14) and using the symmetry of \(\kappa\),

Step 3: \(\lambda\) is additive. Apply Equation (18.121) to \(\left(x\otimes t^{a},\ y\otimes t^{b},\ z\otimes t^{e}\right)\) with \(a+b+e = 0\). The three terms pair \(t^{a+b} = t^{-e}\) with \(t^{e}\), and so on, giving

the three Killing factors being the common value \(\mathcal{K}\) computed in the proof of Theorem A39.6. Some choice of \(x,y,z\) makes \(\mathcal{K} \neq 0\): \(\mathfrak{g} = \comm{\mathfrak{g}}{\mathfrak{g}}\) for a simple algebra, so a nonzero \(w\) with \(\kappa(w,z)\neq0\) — available by nondegeneracy — is a sum of brackets, one of which must pair nontrivially with \(z\). Hence \(\lambda_{p}+\lambda_{q}+\lambda_{r} = 0\) whenever \(p+q+r = 0\); putting \(r = -p-q\) and using Equation (A39.15), \(\lambda_{p+q} = \lambda_{p}+\lambda_{q}\), so \(\lambda_{m} = m\,\lambda_{1}\).

Therefore \(c\left(x\otimes t^{m},y\otimes t^{n}\right) = \lambda_{1}\,m\,\delta_{m+n,0}\,\kappa(x,y)\), which is \(\lambda_{1}\omega_{\kappa}\) by Equation (A39.10).

Derives Lemma A39.11. Abbreviate \(\hat{A} = \hat{A}_{f}\). Expanding the nested commutator,

Taking the ground-state expectation and using \(\hat{H}\ket{0} = E_{0}\ket{0}\) on the last two terms,

Inserting the resolution of the identity \(\sum_{n}\ketbra{n}{n}\) between the operators and using \(\hat{A}^{\dagger} = \hat{A}\), which gives \(\bra{0}\hat{A}\ket{n} = \overline{\bra{n}\hat{A}\ket{0}}\),

which is Equation (A39.18). Every summand is nonnegative because \(E_{n} \ge E_{0}\), so the sum vanishes only if \(\bra{n}\hat{A}\ket{0} = 0\) for every \(n\) with \(E_{n} > E_{0}\), that is only if \(\hat{A}\ket{0}\) has no component outside the ground eigenspace.

Derives Theorem A39.12. The Heisenberg equation \(\ii\hbar\,\pp_{t}\hat{O} = \comm{\hat{O}}{\hat{H}}\) and the continuity equation Equation (A39.16) give

so that, integrating by parts against the compactly supported \(f\),

Therefore

whose ground-state expectation, by Equation (A39.19) and \(\int\dd^{3}y\,f(\vect{y})\,\pp_{k}^{(\vect{x})} \delta^{3}(\vect{x}-\vect{y}) = \left(\pp_{k}f\right)(\vect{x})\), is

the two imaginary units combining to \(-\ii\cdot\ii = 1\). Equating this with Equation (A39.18) gives Equation (A39.20). Since \(\int\abs{\nabla f}^{2} > 0\) for any nonconstant \(f\), the sign of \(S\) follows; and \(S = 0\) makes the right-hand side vanish for every \(f\), so by the equality clause of Lemma A39.11 the vector \(\hat{A}_{f}\ket{0}\) lies in the ground eigenspace for every \(f\), which is the stated conclusion.

The dimensions check: \(\hat{A}_{f}\) carries \(\mathrm{C}\), so the right-hand side of Equation (A39.20) carries \(\mathrm{J}\times\mathrm{C}^{2}\); on the left, \(\hbar S \int\abs{\nabla f}^{2}\dd^{3}x\) carries \(\mathrm{J}\,\mathrm{s}\times \mathrm{C}^{2}/\mathrm{m}/\mathrm{s}\times\mathrm{m}\), the same.